Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
inline int pidorand() { return ((rand() & 32767) << 15) | (rand() & 32767); }
using namespace std;
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
int main() {
  int n = nxt(), m = nxt();
  vector<vector<int>> a(n, vector<int>(m, 0));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      a[i][j] = nxt();
    }
  }
  vector<int> r(n), c(m);
  for (int i = 0; i < n; ++i) {
    r[i] = a[i][0];
  }
  for (int j = 0; j < m; ++j) {
    c[j] = a[0][j] - r[0];
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (a[i][j] != r[i] + c[j]) {
        puts("-1");
        return 0;
      }
    }
  }
  int mr = *min_element((r).begin(), (r).end());
  int mc = *min_element((c).begin(), (c).end());
  int offset = 0;
  if (n <= m) {
    offset = mc;
  } else {
    offset = -mr;
  }
  for (int& x : r) {
    x += offset;
  }
  for (int& x : c) {
    x -= offset;
  }
  for (int x : r) {
    if (x < 0) {
      puts("-1");
      return 0;
    }
  }
  for (int x : c) {
    if (x < 0) {
      puts("-1");
      return 0;
    }
  }
  cout << accumulate((r).begin(), (r).end(), 0) +
              accumulate((c).begin(), (c).end(), 0)
       << "\n";
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < r[i]; ++j) {
      printf("row %d\n", i + 1);
    }
  }
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < c[i]; ++j) {
      printf("col %d\n", i + 1);
    }
  }
  return 0;
}